4 ms·
Look around the room for objects in sight. For each object, take its common name and count how many of the letters are "odd" letters acegikmoqsuwy, then mod 2.
by blackboxlogic 4y ago
Look around the room for objects in sight. For each object, take its common name and count how many of the letters are "odd" letters acegikmoqsuwy, then mod 2. "Window" -> "wiow" -> 4 -> 0. Each word yields a single bit of very slow, pretty good entropy. Don't do this in the same room twice.
- ryukoposting 4y agoI wonder how effective it would be simply to count vowels mod 2. Much faster to calculate, at least for me.
- blackboxlogic 4y agoI suspect there's a strong bias in vowelCount % 2. A quick look at English 100 most common [1] has 1's at 73% and 0's at 27%. That would even out with longer words but I wonder how much. Maybe there's something else that's just as easy with less bias? [1]https://www.englishclub.com/vocabulary/common-words-100.htm https://www.englishclub.com/vocabulary/common-words-100.htm
- ryukoposting 4y agoHm. That makes this puzzle all the more interesting. Maybe it's because I made a Wordle solver app recently, but word statistics are on my mind.
- jhgb 4y agoPerhaps you could do a similar thing with a book. Randomly open it and blindly point to a line in the book. Then do something with that line.